The correct option is 1. f(x) is nowhere continuous.

Explanation:

To determine the continuity of f(x), we need to analyze the behavior of the function as x approaches different points.

Consider x = 0:

When x = 0, the expression becomes f(0) = lim n->infinity [lim t->0 [sin^2(n!pi * 0)/(sin^2(n!pi * 0) * t^2)]].

Since sin(0) = 0, the numerator becomes 0 for all values of n. However, the denominator sin^2(n!pi * 0) = sin^2(0) = 0 for n = 1,2,3,..., which leads to an indeterminate form of 0/0.

By applying L'Hopital's rule, we can differentiate the numerator and denominator with respect to n. Differentiating the numerator gives 0 and differentiating the denominator gives 2sin(n!pi * 0) * n!pi * 0 = 0.

Thus, we have 0/0, which is an indeterminate form. To further evaluate the limit, we can apply L'Hopital's rule again by differentiating the numerator and denominator with respect to n.

Differentiating the numerator again gives 0, and differentiating the denominator again gives 2cos(n!pi * 0) * (n!pi)^2 * 0 = 0.

We can continue this process indefinitely, and at each step, both the numerator and denominator become 0. Therefore, the limit remains indeterminate.

This means that f(0) does not exist, and hence, f(x) is discontinuous at x = 0.

Since the function is discontinuous at x = 0, it cannot be continuous anywhere else. Therefore, option 1 is correct.

Option 2, which states that f(x) has only 2 points of discontinuity, is incorrect. The function is actually discontinuous at every point in its domain, including x = 0.